CHEMISTRY 120A – Inorganic Chemistry – Fall 2011

PROBLEM SET 1

Due Monday, October 17th 2011 by 12:30 pm in Pacific Hall 4100G

Guidelines (Please Read Carefully): Please include this top page in your submission to facilitate the grading of your problem sets and the assignment of grades. Please write clearly (especially your name, student ID and email address). Use additional sheets if necessary. Please bring your UCSD ID with you when you hand in the problem set. Be prepared to sign-out on your problem set to receive credit. Thank you!

Question 1 ________/ 10 Question 9 ________/ 5

Question 2 ________/ 10 Question 10 ________/ 5

Question 3 ________/ 10 Question 11 ________/ 10

Question 4 ________/ 10 Question 12 ________/ 20

Question 5 ________/ 5 Question 13 ________/ 20

Question 6 ________/ 10 Question 14 ________/ 10

Question 7 ________/ 20 Question 15 ________/ 10

Question 8 ________/ 10 Question 16 ________/ 10

Question 17 ________/ 25

NAME: __________ANSWER KEY_______________________ SECTION: ______________________________ PID: ___________________________________ UCSD Email Address: _____________________________

TOTAL

/ 200

 

     

Question 2 (10 points). In general, atomic radii increase significantly as one descends a group in the period table. For example: Li (157 pm) Na (191 pm) K (235) Rb (250) Cs (272). This trend is present between the 3d and 4d transition metals, but not between the 4d and 5d transition metals. For example, there is little difference between the atomic radii of Mo (140 pm) and W (141 pm) or Tc (135) and (Re 137). Using Zeff and the concept of shielding, rationalize why there is little change in radii for 4d and 5d transition metals. Please use the space provided.    Ignoring any relativistic effects, the f-orbitals of the lanthanides poorly shield the valence d-orbitals transition metals in Period 6. Hence, the valence electrons experience a greater Zeff and thus the radii are contracted.                            Question 3 (10 points). Sketch and label the radial distribution functions of 3p and 3s orbitals on the axis provided below. Use the sketch and the concepts of shielding and penetration explain why the Zeff experienced by the 3p orbital is just slightly less than the 3s. Please use the space provided.  The sketch should correctly approximately 3s and 3p radial distribution functions. The 3p orbital is less penetrating and, as a consequence, the 3p electrons are more shielded than 3s. Hence, 3p experience smaller Zeff.

             

 Question 4 (10 points). As a general trend, first ionization energies increase as one moves (from left to right) across the periodic table. Explain why, however, Boron, Aluminum and Gallium have first ionization energies lower than Beryllium, Magnesium, and Calcium, respectively. B, Al and Ga have 1st ionization energies lower than Be, Mg and Ca because the outermost electron occupies a higher energy p orbital. Question 5 (5 points). Rank the following atoms in order of atomic radii (largest to smallest). K, F, Cs, B, Po , Co, V, Ir Cs > K > Ir > V > Co > Po > B >F Question 6 (10 points). Sketch a simple molecular orbital diagram of He2. What is the bond order? Using the MO diagram, rationalize why dihelium is not a stable molecule, but dihydrogen is.  

               

Question 7 (20 points). Trimethyliodosilane is a very potent, electrophilic-silylating agent that reacts readily with nucleophiles to deliver a trimethylsilyl group. On the other hand, trimethylfluorosilane is rather unreactive toward nucleophilic substrates.

I) Write a balanced chemical equation for the reaction between trimethyliodosilane and sodium methoxide.

ISi(CH3)3 + NaOCH3 → H3COSiMe3 + NaI

II) Explain the differences in reactivity between trimethylfluorosilane and trimethyliodosilane toward nucleophiles using Molecular Orbital arguments. a – For both trimethylfluorosilane and trimethyliodosilane, draw a partial MO

diagram between px-orbitals of silicon and the appropriate halide. Be sure to draw a picture of the molecular orbitals formed from the atomic orbitals in your diagrams.

b – Use your diagrams to predict the site of reactivity between nucleophiles and both trimethylfluorosilane and trimethyliodosilane.

c – Use your diagrams to explain the difference in observed reactivity between nucleophiles and both trimethylfluorosilane and trimethyliodosilane.

Use as many pages as necessary.      

     

   

 

                     

3px

2px

Si

F

ψ− = c1χ1 - c2χ2

ψ+ = c1χ1 + c2χ2

χ1

χ2

3px

5pxSi

I

ψ− = c1χ1 - c2χ2

ψ+ = c1χ1 + c2χ2

χ1

χ2

E

Si F

Si F

Si I

Si I

Nucleophiles attack Here (at unoccupied MO of

majority Si parentage)

The Si-F bond is unreactivetoward nucleophiles becausethe Si-F antibonding component(which is largely Si in character)is simply too high in energy.Nucleophile therefore cannotaccess it. This is because ofthe large electronegativitydifference between Si and F.While I is still more electronegative than Si, it is less so than F. Therefore the Si-Ianti-bonding orbital will not be asenergetically destabilized and can beaccessed by nucleophiles.

 

Question 8 (10 points). Name and draw the structure, including any isomers, of:

a. PtCl2(NH3)2:

cis-diamminedichloridoplatinum(II),

trans-diamminedichloridoplatinum(II)

     cis                                       trans

b. K4[Fe(CN)6] potassium hexacyanidoferrate(II)

c. Na[PtBrCl(NO2)(NH3)]: sodium amminebromidochlorido(nitrito-κN)platinate(II)

 

 

 

 

 

PtCl

H3N Cl

NH3

d. [Cu(NH3)5(OH2)]SO4: pentaammineaquacopper(II) sulfate

e. [Fe(κN-NCS)(OH2)5]Br2: pentaaqua(thiocyanato-κN)iron(III) bromide

Question 9 (5 points). Name the following complexes:

a.

b.

c.

d.

e.

a. fac-tricarbonyltriiodidomolybdenum(III) b. mer-triamminetrichloridoiron(III) c. trans-bis(ethylenediamine)diiodidocobalt(III) iodide d. cis-tetraamminedichloridocobalt(III) chloride e. triaquabromidochloridoiodidochromium(III)

Question 10 (5 points). Name the following coordination complexes:

a. CrCl3(NH3)3 d. [CrBr(H2O)5]2+

b. PtCl2(en) e.[CuCl4(NH2CH2CH2NH2)]2-

c. [Pt(ox)2]2- f. Fe(CO)5

a. Triamminetrichloridochromium(III) b. Dichloridoethylenediamineplatinum(II) c. Bis(oxalato)platinate(II) d. Pentaaquabromidochromium(III) e. Tetrachloridoethylenediaminecuprate(II) f. Tetrahydroxidoferrate(III)

Question 11 (10 points). To the right of the name, draw the structures of the following compounds:

a. Potassium hexabromidoplatinate(IV)

b. Potassium diamminetetrachloridocobaltate(III)

c. Tris(ethylenediamine)copper(II) sulfate

d. Hexacarbonylmanganese(I) perchlorate

e. Ammonium tetrachloridoruthenate(III)

Question 12 (20 points). Below and on a separate page, draw all possible isomers of:

a. [RhH(C≡CR)2(PMe3)3] b. [PtBrCl(NH3)(py)] c. [IrCl2(NH3)3]+ (trigonal bipyramidal) d. [FeBr2(py)3] (square pyramidal) e. [Ru(dmpe)3]2+

(dmpe = (CH3)2PCH2CH2P(CH3)2; bis-dimethylphosphinoethane)

a. [RhH(C≡CR)2(PMe3)3] (3 isomers)

fac-[RhH(C≡CR)2(PMe3)3] mer-trans-[RhH(C≡CR)2(PMe3)3]

RuCl

Cl Cl

Cl

NH4

mer-cis-[RhH(C≡CR)2(PMe3)3]

b. [PtBrClNH3(py)]

c. [Ir(Cl)2(NH3)3]+ (trigonal bipyramidal)

d. [Fe(Br)2(py)3] (square pyramidal)

 

 

e. [Ru(dmpe)3]2+

Δ-[Ru(dmpe)3]2+ Λ-[Ru(dmpe)3]2+

Question 13 (20 points). Determine the point group of the following molecules.

   

Question 14 (10 points). Perform the symmetry operations below on the dxz orbital shown and display these results pictorially. In the given Cartesian coordinate system, what is the new identity of the d-orbital following the symmetry operation? In C2v symmetry, what Mulliken symbols would be assigned to these new orbitals?   a) i b) C4 (along z-azis) c) C4 (along x-axis) d) S2 (along z-axis)    

 

   

a) B1 b) B2 c) A2 d) B1  Question 15 (10 points). Perform the symmetry operations of the D2h point group on a pz orbital and display these results pictorially. Please be sure to indicate the character that is returned when the symmetry operation is applied. Based on this unique set of characters, what is the irreducible representation of this orbital? Are any other orbitals degenerate with this orbital in D2h point symmetry? No degenerate orbitals (B1u).

 

Question 16 (10 points). What new point group is obtained by removing or adding the indicated symmetry operation from each of the following point groups? a) C3 plus i ->S6 b) C3v plus i ->D3d c) C5v plus σh ->D5h d) S6 minus i ->C3

Descriptions: a) You cannot add an inversion center to the C3 point group without in the process creating a S6 symmetry element. Remember, the inversion center requires that for every (x,y,z) point within a molecule there is a corresponding (-x,-y,-z) point. The S6 point group contains these added symmetry elements. b) You cannot add an inversion center to the C3v point group without in the process creating a S6 symmetry element. However, in C3v symmetry there are also 3 σv’s, and consequently adding inversion symmetry also creates 3 perpendicular C2 axis. The D3d point group contains these added symmetry elements. c) As a consequence of the 5 σv’s, in the C5v point group, adding σh will generate 5 perpendicular C2’s. The D3h point group contains these added symmetry elements. d) In removing inversion symmetry from the S6 point group, the S6 symmetry element is also removed. The remaining symmetry elements are those of the C3 point group. Question 17 (25 points). Consider the [BF2]+ cation below. This molecule possesses a trivalent central boron atom. Fluorine and boron can form especially strong out-of-plane π-bonds. Using this information, complete the following (this question contains parts a-d). a) Determine the point group of [BF2]+ and, using the coordinate system above, assign the atomic orbitals on boron to their respective Mulliken Symbols.

Point Group = C2v

B

Fa Fbx

z

y

b) Generate the σ-bonding framework for [BF2]+: Determine the irreducible representations for the fluorine group orbitals in the (SALCs) σ-direction. You can assume that fluorine uses only its s-orbitals for σ-bonding. To receive full credit, clearly show all your work. Use an extra page if necessary. Character summary for s-bonding SALC’s (made up of fluorine s-obitals). Note change in B1 vs. B2 due to orientation.

A1

s pypx pz

A1B1 B2

A1

B1

E C2 σv(xz) σ'v(yz)

Γ

C2v

2 0 02

1 1 1 1

1 -1 1 -1

ψ(A1) = 1/21/2[Fa+Fb] ψ(B1) = 1/21/2[Fa-Fb]

c) Sketch a Molecular Orbital Diagram for the σ-framework of [BF2]+ and draw all bonding, non-bonding and anti-bonding molecular orbitals. Also Indicate the symmetry of each molecular orbital. Be sure to draw pictures of the molecular orbitals (i.e. AO’s + SALCs) to accompany their placement in the Molecular Orbital diagram.

2sA1

2py2px 2pz

A1B2B1

A1

B1

2s

1a1

1b1

2a1

3a1

2b1

1a1

1b2

2a1

1b1

3a1

2b1

[B]+

2F

1b2

Fluorine 2s orbitals lower in Ethan B 2s and 2p orbitals. Each F uses only one e- in σ-bonding

BONDING (σ)

ANTI-BONDING (σ∗)

NON-BONDING and Empty

A1 orbital of B 2pz parentagemixes with A1 B 2s orbital,thereby lowering its energya bit. This orbital will also mixslightly with A1 SALC in abonding manner, but the overallcharacter is non-bonding.

d) Generate the π-bonding framework of [BF2]+. Determine the SALCs for the out-of-plane p-orbitals on fluorine and pair them with the appropriate AO’s on boron. Draw an MO diagram that shows all π-interactions between boron and the fluorine atoms in [BF2]+ (there may be more than one). If more than one i) determine which π-interaction is more stabilizing and indicate such pictorially on the diagram ii) write 2-3 sentences explaining your choice in your own words (speak in the language of molecular symmetry). Again, be sure to draw pictures of the molecular orbitals (i.e. AO’s + SALCs) to accompany their placement in the Molecular Orbital diagram. π-bonding occurs in the direction perpendicular to σ-bonding. In the bent form of [BF2]+, in the coordinate and orientation shown, both the 1b2 (boron py orbital) and the 2a1 (boron pz orbital) are non-bonding, empty and are perpendicular to the σ-bonding vector (yz-plane). These orbitals can therefore be used for π-bonding with the appropriate lone pairs on the F atoms. In this molecule, p-bonding will arise from the donation of an electron pair from F to a suitable empty orbital on B (i.e. the empty 1b2(py) and 2a1(pz) orbitals). STEP 1 - Generate SALCS of F lone pairs to interact with 1b2 MO (boron py):      

 

             STEP 2 - Generate SALCS of F lone pairs to interact wit 2a1 MO (boron pz):                      

x

z

y A2

B2

E C2 σv(xz) σ'v(yz)

Γ

C2v

2 0 0-2

1 1 -1 -1

1 -1 1-1Fa Fb

py orbitals of Fa and Fb

ψ(B2) = 1/21/2[Fa+Fb] ψ(A2) = 1/21/2[Fa-Fb]

SALCs =

x

z

yA1

B1

E C2 σv(xz) σ'v(yz)

Γ

C2v

2 0 02

1 1 1 1

1 -1 1 -1FaFb

pz orbitals of Fa and Fb

ψ(A1) = 1/21/2[Fa+Fb] ψ(B1) = 1/21/2[Fa-Fb]

SALCs =

 STEP 3 – Add π-symmetry fluorine SALCs to σ-bonding MO diagram, complete interactions:

 

2sA1

2py2px 2pz

A1B2B1

A1

B1

2s

1a1

1b1

2a1

3a1

2b1

1b2

[B]+

2F

A1

B1

A2

B2

1b2(π)

2b2(π∗)

2a1(π)

3a1(π∗)

1a2 and 2b1 (non-boning/lone pairs

are just the SALCs)

π-bonding framework

1b2(π)

2a1(π)

1b2(π∗)

2a1(π∗)

Much better overlap between the B2 SALCthe boron empty py orbital (b2 MO) thanthere is in the A1 π-type salc and the 2a1 MO.Therefore this will be a stronger, more stable (also lower energy interaction). Thecorresponding b2 π* orbital will also behigher in energy than the a1 π* orbital .

2F 2py and 2pz